TITLE: Why should the real numbers interact with number fields and Galois theory?
QUESTION [6 upvotes]: When I wanted to show $\mathbb{Q}(2^{1/3})/\mathbb{Q}$ is not Galois, my argument was it's a real field, while the normal closure clearly contains complex numbers. However I'm unsatisfied with this argument, suppose someone who never met $\mathbb{R}$ asks this, he just innocently takes $x^3-2$ and wonders if adjoining one root gives a Galois extension. How would he come up with the $\mathbb{R}$ argument? It feels unmotivated- it seems he'd need to say something like the algebraic closure of $\mathbb{Q}$ has an order two automorphism, and understand its interaction with the roots of $x^2 - 3$.
** There is an argument that avoids $\mathbb{R}$, namely that $\phi(3)=2$ so this divides the splitting field index, but it's a different argument and requires $\phi(n) \nmid n$.
Thus my question is:

Are there more arguments for why $\mathbb{Q}(2^{1/3})/\mathbb{Q}$ is not Galois
Is there a better motivation to how and why $\mathbb{R}$ is helpful, and how one naturally arrives at it?
Are there more striking examples of $\mathbb{R}$ playing an 'algebraic' role? (I'm familiar with the geometry of numbers, but that's not the type of examples I'm looking for, there it plays an analytic role).

REPLY [3 votes]: We can show $K = \mathbb{Q}(\sqrt[3]{2})$ is not Galois using only computations over $\mathbb{Q}$, as follows:
We want to show the only root of $f=T^3 - 2$ in $K$ is $\alpha=\sqrt[3]{2}$. Suppose that $\beta \in K$ is a root of $f$, and write $\beta = x + y\alpha + z\alpha^2$. Compute the coefficients of $\beta^3 - 2$ w.r.t the basis $(1,\alpha,\alpha^2)$, by hand or like this in SageMath:
sage: K.<a> = NumberField(x^3 - 2)
sage: A.<x,y,z> = PolynomialRing(K)
sage: v = sum(vector(c)*m for (c,m) in (x+y*a+z*a^2)^3 - 2)
sage: show(v)

$$\left(x^{3} + 2 y^{3} + 12 x y z + 4 z^{3} - 2,\,3 x^{2} y + 6 y^{2} z + 6 x z^{2},\,3 x y^{2} + 3 x^{2} z + 6 y z^{2}\right)$$
These three polynomials define an ideal $I$ in the ring $\mathbb{Q}[x,y,z]$:
sage: B.<x,y,z> = PolynomialRing(QQ, order='lex')
sage: I = B.ideal(list(v))

Computing a Gröbner basis of $I$ w.r.t lexicographic ordering shows that $V(I)$ is finite:
sage: I.groebner_basis()
[x^3 + 2*y^3 + 3240/7*z^9 - 216*z^6 - 26/7*z^3 - 2,
 x*y - 270/7*z^8 + 18*z^5 + 9/14*z^2,
 x*z - 54*y^2*z^6 + 9*y^2*z^3,
 y^4 - 108*y*z^6 - 2*y*z^3 - y,
 y^3*z + 27/7*z^7 - 2*z^4 + 1/28*z,
 y*z^7 + 1/108*y*z,
 z^10 - 1/2*z^7 + 1/108*z^4 - 1/216*z]

The last element gives the equation $216z^{10} - 108z^7 + 2z^4 - z = 0$ which, by the rational root theorem, has $z=0$ as its only root in $\mathbb{Q}$. The fourth element then yields $y^4 - y = 0$, hence for $y \in \mathbb{Q}$ we get $y=0$ or $y=1$. If $y=0$ then the first element yields $x^3 - 2 = 0$ which is a contradiction for $x \in \mathbb{Q}$. So $y=1$, and the second element shows $x=0$. In summary $(x,y,z) = (0,1,0)$, which means $\beta=\alpha$, so we are done.
In general there are algorithms to compute a Gröbner basis, the radical of a zero-dimensional ideal, the intersections with the univariate polynomial rings, and rational roots of univariate polynomials. This may be not the most efficient way, but it works and uses only arithmetic in $\mathbb{Q}$.